For me the definition of amenability of an at most countable discrete group (with counting measure) is existence of a Folner sequence. Assuming this, why is every countable discrete abelian group amenable? What is the Folner sequence that does the job?
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asked Jun 12, 2010 at 19:15
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$\begingroup$ This is a natural and reasonable question, but isn't it essentially covered by the discussion at mathoverflow.net/questions/12169/… ? Perhaps this should be closed as a duplicate? $\endgroup$– Yemon ChoiJun 12, 2010 at 19:54
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$\begingroup$ Hm, so the difficulty comes from the fact that it's not necessarily finitely generated then? $\endgroup$– Harry AltmanJun 12, 2010 at 20:01
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$\begingroup$ I had a look at the topic in the link before, but to me it wasn't clear how to pick the generators in Tom Church's answer there in a way that one can control the size of the "pyramid" sets appearing in the Folner sequence.. Yes, the difficult case is when the group is not necessarily finitely generated. $\endgroup$– Kestutis CesnaviciusJun 12, 2010 at 20:46
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The direct limit $G = \bigcup_n G_n$ of a nested sequence of countable amenable groups $G_n$ is still amenable, since every finite set $S$ in $G$ will lie in one of the $G_n$ and thus there must exist some finite set $F_S$ which is not shifted very much by $S$. Since there are only a countable number of $S$, one can diagonalise and obtain a Folner sequence for $G$.
Since every countable abelian group is the direct limit of finitely generated abelian groups, which have polynomial growth and are thus amenable, every countable abelian group is amenable.
An instructive example is the free group $\bigcup_n Z^n$ on countably many generators. Here, the sets $\{-N_n,\ldots,N_n\}^n$ will form a Folner sequence if $N_n$ grows sufficiently rapidly in n.
I have some notes on amenability that cover these topics at
http://terrytao.wordpress.com/2009/04/14/some-notes-on-amenability/
answered Jun 13, 2010 at 5:48